【BZOJ2754】[SCOI2012]喵星球上的点名

【BZOJ2754】[SCOI2012]喵星球上的点名

题面

bzoj

洛谷

题解

这题有各种神仙做法啊，什么暴力\(AC\)自动机、\(SAM\)等等五花八门

我这个蒟蒻在这里提供一种复杂度正确且常数小的做法。

根据后缀数组经典套路，

我们用一个未出现过的字符将所有串连接起来求一边\(SA\)（不算询问串）

然后因为我们现在已经将所有后缀排好序了

而询问串要有贡献一定是一个后缀的前缀

所以一个区间能产生贡献的后缀是排名连续的一段

这样这段区间的左右端点可以二分出来

我们再对于每个位置记一下它是哪一个人的

这样第一问就被转化成了区间内有多少种颜色，直接上莫队在排名上跑即可

到了这一步

第二问就很好做了

我们考虑差分，每新遇到一类数，我们加上剩余询问个数的贡献

每去掉一类数，减去剩余询问个数的贡献即可

代码

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
inline int gi() {
	register int data = 0, w = 1;
	register char ch = 0;
	while (!isdigit(ch) && ch != '-') ch = getchar();
	if (ch == '-') w = -1, ch = getchar();
	while (isdigit(ch)) data = 10 * data + ch - '0', ch = getchar();
	return w * data;
}
const int MAX_N = 2e5 + 5;
int N, n, m, a[MAX_N], id[MAX_N], lim = 1e4 + 5;
int sa[MAX_N];
void GetSA() {
#define cmp(i, j, k) (y[i] == y[j] && y[i + k] == y[j + k])
    static int x[MAX_N], y[MAX_N], bln[MAX_N];
    int M = lim;
    for (int i = 1; i <= N; i++) bln[x[i] = a[i]]++;
    for (int i = 1; i <= M; i++) bln[i] += bln[i - 1];
    for (int i = N; i >= 1; i--) sa[bln[x[i]]--] = i;
    for (int k = 1; k <= N; k <<= 1) {
        int p = 0;
        for (int i = 0; i <= M; i++) y[i] = 0;
        for (int i = N - k + 1; i <= N; i++) y[++p] = i;
        for (int i = 1; i <= N; i++) if (sa[i] > k) y[++p] = sa[i] - k;
        for (int i = 0; i <= M; i++) bln[i] = 0;
        for (int i = 1; i <= N; i++) bln[x[y[i]]]++;
        for (int i = 1; i <= M; i++) bln[i] += bln[i - 1];
        for (int i = N; i >= 1; i--) sa[bln[x[y[i]]]--] = y[i];
        swap(x, y); x[sa[1]] = p = 1;
        for (int i = 2; i <= N; i++) x[sa[i]] = cmp(sa[i], sa[i - 1], k) ? p : ++p;
        if (p >= N) break;
        M = p;
    }
}
const int LEN = 400;
int cnt[MAX_N], bln[MAX_N];
struct Query { int l, r, id; } q[MAX_N]; int q_cnt = 0;
inline bool operator < (const Query &a, const Query &b) {
	if (bln[a.l] ^ bln[b.l]) return a.r < b.r;
	else return (bln[a.l] & 1) ? a.r < b.r : a.r > b.r;
}
int A1, ans1[MAX_N], ans2[MAX_N]; 

void add(int x, int pos) {
	cnt[id[sa[x]]]++;
	if (cnt[id[sa[x]]] == 1) ++A1, ans2[id[sa[x]]] += q_cnt - pos + 1;
}
void del(int x, int pos) {
	cnt[id[sa[x]]]--;
	if (cnt[id[sa[x]]] == 0) --A1, ans2[id[sa[x]]] -= q_cnt - pos + 1;
}
int main () {
#ifndef ONLINE_JUDGE
	freopen("cpp.in", "r", stdin);
#endif
	n = gi(), m = gi();
	for (int i = 1, L; i <= n; i++) {
		L = gi(); for (int j = 1; j <= L; j++) a[++N] = gi(), id[N] = i; a[++N] = ++lim;
		L = gi(); for (int j = 1; j <= L; j++) a[++N] = gi(), id[N] = i; a[++N] = ++lim;
	}
	GetSA();
	for (int i = 1; i <= m; i++) {
		int L = 1, R = N;
		for (int len = gi(), j = 1; j <= len; ++j) {
			int x = gi(), l = L, r = R;
			while (l <= r) {
                int mid = (l + r) >> 1;
				if (a[sa[mid] + j - 1] < x) l = mid + 1;
				else r = mid - 1;
			}
			int tmp = l; l = L, r = R;
			while (l <= r) {
				int mid = (l + r) >> 1;
				if (a[sa[mid] + j - 1] <= x) l = mid + 1;
				else r = mid - 1;
			}
			L = tmp, R = r;
		}
		if (L <= R) q[++q_cnt] = (Query){L, R, i};
	}
	for (int i = 1; i <= N; i++) bln[i] = (i - 1) / LEN + 1;
	sort(&q[1], &q[q_cnt + 1]);
	for (int ql = 1, qr = 0, i = 1; i <= q_cnt; i++) {
		while (ql < q[i].l) del(ql, i), ++ql;
	    while (ql > q[i].l) --ql, add(ql, i);
	    while (qr < q[i].r) ++qr, add(qr, i);
	    while (qr > q[i].r) del(qr, i), --qr;
		ans1[q[i].id] = A1;
	}
	for (int i = 1; i <= m; i++) printf("%d\n", ans1[i]);
	for (int i = 1; i <= n; i++) printf("%d ", ans2[i]);
	return 0;
}

跑得还是挺快的，在洛谷跑了\(Rank4\)